IJMMS 31:3 (2002) 167–175
PII. S0161171202109288
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
CR-SUBMANIFOLDS OF A NEARLY TRANS-SASAKIAN MANIFOLD
FALLEH R. AL-SOLAMY
Received 26 September 2001
This paper considers the study of CR-submanifolds of a nearly trans-Sasakian manifold,
generalizing the results of trans-Sasakian manifolds and thus those of Sasakian manifolds.
2000 Mathematics Subject Classification: 53C40.
1. Introduction. In 1978, Bejancu introduced the notion of CR-submanifold of a
Kaehler manifold [1]. Since then several papers on CR-submanifolds of Kaehler manifold have been published. On the other hand, CR-submanifolds of a Sasakian manifold
have been studied by Kobayashi [6], Shahid et al. [10], Yano and Kon [11], and others.
Bejancu and Papaghuic [2] studied CR-submanifolds of a Kenmotsu manifold. In 1985,
Oubina introduced a new class of almost contact metric manifold known as transSasakian manifold [7]. This class contains α-Sasakian and β-Kenmotsu manifold [5].
Geometry of CR-submanifold of trans-Sasakian manifold was studied by Shahid [8, 9].
A nearly trans-Sasakian manifold [4] is a more general concept.
The results of this paper are the generalization of the results obtained earlier by
several authors, namely [6, 8, 9] and others.
2. Preliminaries. Let M be an n-dimensional almost contact metric manifold with
an almost contact metric structure (φ, ξ, η, g) satisfying [3]
φ2 X = −X + η(X)ξ,
η(ξ) = 1, φ ◦ ξ = 0,
g(φX, φY ) = g(X, Y ) − η(X)η(Y ),
g(X, ξ) = η(X),
(2.1)
where X and Y are vector fields tangent to M.
An almost contact metric structure (φ, ξ, η, g) on M is called trans-Sasakian if [7]
X φ (Y ) = α g(X, Y )ξ − η(Y )X + β g(φX, Y )ξ − η(Y )φX ,
(2.2)
where α and β are nonzero constants, denotes the Riemannian connection of g on
M, and we say that the trans-Sasakian structure is of type (α, β).
Further, an almost contact metric manifold M(φ, ξ, η, g) is called nearly transSasakian if [4]
X φ (Y ) + Y φ (X) = α 2g(X, Y )ξ − η(Y )X − η(X)Y
(2.3)
− β η(Y )φX + η(X)φY .
It is clear that any trans-Sasakian manifold, and thus any Sasakian manifold, satisfies the above relation.
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FALLEH R. AL-SOLAMY
Let M be an m-dimensional isometrically immersed submanifold of a nearly transSasakian manifold M and denote by the same g the Riemannian metric tensor field
induced on M from that of M.
Definition 2.1. An m-dimensional Riemannian submanifold M of a nearly transSasakian manifold M is called a CR-submanifold if ξ is tangent to M and there exists
a differentiable distribution D : x ∈ M → Dx ⊂ Tx M such that
(1) the distribution Dx is invariant under φ, that is, φDx ⊂ Dx for each x ∈ M;
(2) the complementary orthogonal distribution D ⊥ : x ∈ M → Dx⊥ ⊂ Tx M of D is
anti-invariant under φ, that is, φDx⊥ ⊂ Tx⊥ M for all x ∈ M, where Tx M and Tx⊥ M
are the tangent space and the normal space of M at x, respectively.
If dim Dx⊥ = 0 (resp., dim Dx = 0), then the CR-submanifold is called an invariant
(resp., anti-invariant) submanifold. The distribution D (resp., D ⊥ ) is called the horizontal (resp., vertical) distribution. Also, the pair (D, D ⊥ ) is called ξ-horizontal (resp.,
vertical) if ξx ∈ Dx (resp., ξx ∈ Dx⊥ ) [6].
For any vector field X tangent to M, we put [6]
X = P X + QX,
(2.4)
⊥
where P X and QX belong to the distribution D and D .
For any vector field N normal to M, we put [6]
φN = BN + CN,
(2.5)
where BN (resp., CN) denotes the tangential (resp., normal) component of φN.
Let (resp., ) be the covariant differentiation with respect to the Levi-Civita connection on M (resp., M). The Gauss and Weingarten formulas for M are respectively
given by
X Y = X Y + h(X, Y );
X N = −AN X + ⊥
(2.6)
X N,
for X, Y ∈ T M and N ∈ T ⊥ M, where h (resp., A) is the second fundamental form (resp.,
tensor) of M in M, and ⊥ denotes the normal connection. Moreover, we have
(2.7)
g h(X, Y ), N = g AN X, Y .
3. Some basic lemmas. First we prove the following lemma.
Lemma 3.1. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M.
Then
P X φP Y + P Y φP X − P AφQX Y − P AφQY X
= φP X Y + φP Y X + 2αg(X, Y )P ξ − αη(Y )P X
− αη(X)P Y − βη(Y )φP X + βη(X)φP Y ,
Q X φP Y + Q Y φP X − Q AφQX Y − Q AφQY X
= 2Bh(X, Y ) − αη(Y )QX − αη(X)QY + 2αg(X, Y )Qξ,
⊥
h(X, φP Y ) + h(Y , φP X) + ⊥
X φQY + Y φQX
= φQ Y X + φQ X Y + 2Ch(X, Y ) − βη(Y )φQX − βη(X)φQY
for X, Y ∈ T M.
(3.1)
(3.2)
(3.3)
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CR-SUBMANIFOLDS OF A NEARLY TRANS-SASAKIAN . . .
Proof. From the definition of the nearly trans-Sasakian manifold and using (2.4),
(2.5), and (2.6), we get
X φP Y + h(X, φP Y ) − AφQY X + ⊥
X φQY − φ X Y + h(X, Y )
+ Y φP X + h(Y , φP X) − AφQX Y + ⊥
Y φQX − φ Y X + h(X, Y )
= α 2g(X, Y )ξ − η(Y )X − η(X)Y − β η(Y )φX + η(X)φY
(3.4)
for any X, Y ∈ T M.
Now using (2.4) and equaling horizontal, vertical, and normal components in (3.4),
we get the result.
Lemma 3.2. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M.
Then
2 X φ (Y ) = X φY − Y φX + h(X, φY ) − h(Y , φX) − φ[X, Y ]
+ α 2g(X, Y )ξ − η(Y )X − η(X)Y − β η(Y )φX + η(X)φY
(3.5)
for any X, Y ∈ D.
Proof. By Gauss formula (2.6), we get
X φY − Y φX = X φY + h(X, φY ) − Y φX − h(Y , φX).
(3.6)
Also, we have
X φY − Y φX = X φ (Y ) − Y φ (X) + φ[X, Y ].
(3.7)
From (3.6) and (3.7), we get
X φ (Y ) − Y φ (X) = X φY + h(X, φY ) − Y φX − h(Y , φX) − φ[X, Y ].
(3.8)
Also for nearly trans-Sasakian manifolds, we have
X φ (Y ) + Y φ (X) = α 2g(X, Y )ξ − η(Y )X − η(X)Y − β η(Y )φX + η(X)φY .
(3.9)
Combining (3.8) and (3.9), the lemma follows.
In particular, we have the following corollary.
Corollary 3.3. Let M be a ξ-vertical CR-submanifold of a nearly trans-Sasakian
manifold, then
2 X φ (Y ) = X φY − Y φX + h(X, φY ) − h(Y , φX) − φ[X, Y ] + 2αg(X, Y )ξ
(3.10)
for any X, Y ∈ D.
Similarly, by Weingarten formula (2.6), we get the following lemma.
170
FALLEH R. AL-SOLAMY
Lemma 3.4. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M, then
⊥
2 Y φ (Z) = AφY Z − AφZ Y + ⊥
Y φZ − Z φY − φ[Y , Z]
(3.11)
+ α 2g(Y , Z)ξ − η(Y )Z − η(Z)Y − β η(Y )φZ + η(Z)φY
for any Y , Z ∈ D ⊥ .
Corollary 3.5. Let M be a ξ-horizontal CR-submanifold of a nearly trans-Sasakian
manifold, then
⊥
2 Y φ (Z) = AφY Z − AφZ Y + ⊥
Y φZ − Z φY − φ[Y , Z] + 2αg(Y , Z)ξ
(3.12)
for any Y , Z ∈ D ⊥ .
Lemma 3.6. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M, then
2 X φ (Y ) = α 2g(X, Y )ξ − η(Y )X − η(X)Y − β η(Y )φX + η(X)φY
(3.13)
− AφY X + ⊥
X φY − Y φX − h(Y , φX) − φ[X, Y ]
for any X ∈ D and Y ∈ D ⊥ .
4. Parallel distributions
Definition 4.1. The horizontal (resp., vertical) distribution D (resp., D ⊥ ) is said to
be parallel [1] with respect to the connection on M if X Y ∈ D (resp., Z W ∈ D ⊥ )
for any vector field X, Y ∈ D (resp., W , Z ∈ D ⊥ ).
Now we prove the following proposition.
Proposition 4.2. Let M be a ξ-vertical CR-submanifold of a nearly trans-Sasakian
manifold M. If the horizontal distribution D is parallel, then
h(X, φY ) = h(Y , φX)
(4.1)
for all X, Y ∈ D.
Proof. Using parallelism of horizontal distribution D, we have
X φY ∈ D,
Y φX ∈ D
for any X, Y ∈ D.
(4.2)
Thus using the fact that QX = QY = 0 for Y ∈ D, (3.2) gives
Bh(X, Y ) = g(X, Y )Qξ
for any X, Y ∈ D.
(4.3)
Also, since
φh(X, Y ) = Bh(X, Y ) + Ch(X, Y ),
(4.4)
then
φh(X, Y ) = g(X, Y )Qξ + Ch(X, Y )
for any X, Y ∈ D.
(4.5)
Next from (3.3), we have
h(X, φY ) + h(Y , φX) = 2Ch(X, Y ) = 2φh(X, Y ) − 2g(X, Y )Qξ,
(4.6)
CR-SUBMANIFOLDS OF A NEARLY TRANS-SASAKIAN . . .
171
for any X, Y ∈ D. Putting X = φX ∈ D in (4.6), we get
h(φX, φY ) + h Y , φ2 X = 2φh(φX, Y ) − 2g(φX, Y )Qξ
(4.7)
h(φX, φY ) − h(Y , X) = 2φh(φX, Y ) − 2g(φX, Y )Qξ.
(4.8)
or
Similarly, putting Y = φY ∈ D in (4.6), we get
h(φY , φX) − h(X, Y ) = 2φh(X, φY ) − 2g(X, φY )Qξ.
(4.9)
Hence from (4.8) and (4.9), we have
φh(X, φY ) − φh(Y , φX) = g(X, φY )Qξ − g(φX, Y )Qξ.
(4.10)
Operating φ on both sides of (4.10) and using φξ = 0, we get
h(X, φY ) = h(Y , φX)
(4.11)
for all X, Y ∈ D.
Now, for the distribution D ⊥ , we prove the following proposition.
Proposition 4.3. Let M be a ξ-vertical CR-submanifold of a nearly trans-Sasakian
manifold M. If the distribution D ⊥ is parallel with respect to the connection on M, then
AφY Z + AφZ Y ∈ D ⊥
for any Y , Z ∈ D ⊥ .
(4.12)
Proof. Let Y , Z ∈ D ⊥ , then using Gauss and Weingarten formula (2.6), we obtain
⊥
− AφZ Y + ⊥
Y φZ − AφY Z + Z φY
= φ Y Z + φ Z Y + 2φh(Y , Z)
+ α 2g(X, Y )ξ − η(Y )Z − η(Z)Y − β η(Y )φZ + η(Z)φY
(4.13)
for any Y , Z ∈ D ⊥ . Taking inner product with X ∈ D in (4.13), we get
g AφY Z, X + g AφZ Y , X = g Y Z, φX + g Z Y , φX .
(4.14)
If the distribution D ⊥ is parallel, then Y Z ∈ D ⊥ and Z Y ∈ D ⊥ for any Y , Z ∈ D ⊥ .
So from (4.14) we get
g AφY Z, X + g AφZ Y , X = 0
or
g AφY Z + AφZ Y , X = 0
(4.15)
which is equivalent to
AφY Z + AφZ Y ∈ D ⊥
for any Y , Z ∈ D ⊥ ,
(4.16)
and this completes the proof.
Definition 4.4 [6]. A CR-submanifold is said to be mixed totally geodesic if
h(X, Z) = 0 for all X ∈ D and Z ∈ D ⊥ .
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FALLEH R. AL-SOLAMY
The following lemma is an easy consequence of (2.7).
Lemma 4.5. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M.
Then M is mixed totally geodesic if and only if AN X ∈ D for all X ∈ D.
Definition 4.6 [6]. A normal vector field N ≠ 0 is called D-parallel normal section
if ⊥
X N = 0 for all X ∈ D.
Now we have the following proposition.
Proposition 4.7. Let M be a mixed totally geodesic ξ-vertical CR-submanifold of a
nearly trans-Sasakian manifold M. Then the normal section N ∈ φD ⊥ is D-parallel if
and only if X φN ∈ D for all X ∈ D.
Proof. Let N ∈ φD ⊥ . Then from (3.2) we have
Q Y φX = 0
for any X ∈ D, Y ∈ D ⊥ .
(4.17)
In particular, we have Q(Y X) = 0. By using it in (3.3), we get
⊥
X φQY = φQ X Y
or
⊥
X N = −φQ X φN.
(4.18)
Thus, if the normal section N ≠ 0 is D-parallel, then using Definition 4.6 and (4.18),
we get
(4.19)
φQ X φN = 0
which is equivalent to X φN ∈ D for all X ∈ D. The converse part easily follows
from (4.18).
5. Integrability conditions of distributions. First we calculate the Nijenhuis tensor
Nφ (X, Y ) on a nearly trans-Sasakian manifold M. For this, first we prove the following
lemma.
Lemma 5.1. Let M be a nearly trans-Sasakian manifold, then
φX φ (Y ) = 2αg(φX, Y )ξ − η(Y )φX + βη(Y )X − βη(X)η(Y )ξ
− η(X)Y ξ + φ Y φ (X) + η Y X ξ
(5.1)
for any X, Y ∈ T M.
Proof. From the definition of nearly trans-Sasakian manifold M, we have
φX φ (Y ) = 2αg(φX, Y )ξ − η(Y )φX + βη(Y )X
− βη(Y )η(X)ξ − Y φ (φX).
(5.2)
Also, we have
Y φ (φX) = Y φ2 X − φY φX
= Y φ2 X − φY φX + φ φY X − φ φY X
= −Y X + η(X)Y ξ − φ Y φX − φY X − φ φY X
= −Y X + η(X)Y ξ − φ Y φ (X) − Y X − η Y X ξ.
(5.3)
CR-SUBMANIFOLDS OF A NEARLY TRANS-SASAKIAN . . .
Using (5.3) in (5.2), we get
φX φ (Y ) = 2αg(φX, Y )ξ − η(Y )φX + βη(Y )X − βη(X)η(Y )ξ
− η(X)Y ξ + φ Y φ (X) + η Y X ξ
173
(5.4)
for any X, Y ∈ T M, which completes the proof of the lemma.
On a nearly trans-Sasakian manifold M, Nijenhuis tensor is given by
Nφ (X, Y ) = φX φ (Y ) − φY φ (X) − φ X φ (Y ) + φ Y φ (X)
(5.5)
for any X, Y ∈ T M.
From (5.1) and (5.5), we get
Nφ (X, Y ) = 4αg(φX, Y )ξ − η(Y )φX + η(X)φY + βη(Y )X
− βη(X)Y − η(X)Y ξ + η(Y )X ξ + η Y X ξ
− η X Y ξ + 2φ Y φ (X) − 2φ X φ (Y ).
(5.6)
Thus using (2.3) in the above equation and after some calculations, we obtain
Nφ (X, Y ) = 4αg(φX, Y )ξ + (2α − 1)η(Y )φX + (2α + 1)η(X)φY − βη(Y )X
− 3βη(X)Y + 4βη(X)η(Y )ξ − η(X)Y ξ + η(Y )X ξ
+ η Y X ξ − η X Y ξ + 4φ Y φ (X)
(5.7)
for any X, Y ∈ T M.
Now we prove the following proposition.
Proposition 5.2. Let M be a ξ-vertical CR-submanifold of a nearly trans-Sasakian
manifold M. Then, the distribution D is integrable if the following conditions are satisfied:
S(X, Z) ∈ D,
h(X, φZ) = h(φX, Z)
(5.8)
for any X, Z ∈ D.
Proof. The torsion tensor S(X, Y ) of the almost contact structure (φ, ξ, η, g) is
given by
S(X, Y ) = Nφ (X, Y ) + 2dη(X, Y )ξ = Nφ (X, Y ) + 2g(φX, Y )ξ.
(5.9)
Thus, we have
S(X, Y ) = [φX, φY ] − φ[φX, Y ] − φ[X, φY ] + 2g(φX, Y )ξ
(5.10)
for any X, Y ∈ T M.
Suppose that the distribution D is integrable. So for X, Y ∈ D, Q[X, Y ] = 0 and
η([X, Y ]) = 0 as ξ ∈ D ⊥ .
If S(X, Y ) ∈ D, then from (5.7) and (5.9) we have
2(2α+1)g(φX, Y )ξ +η [X, Y ] ξ +4 φY φX +φh(Y , φX)+QY X +h(X, Y ) ∈ D
(5.11)
174
FALLEH R. AL-SOLAMY
or
2(2α+1)g(φX, Y )Qξ+η [X, Y ] Qξ+4 φQY φX+φh(Y , φX)+QY X+h(X, Y ) = 0
(5.12)
for any X, Y ∈ D.
Replacing Y by φZ for Z ∈ D in the above equation, we get
2(2α + 1)g(φX, φZ)Qξ + 4 φQ φY φX + φh(φZ, φX) + Q φZ X + h(X, φZ) = 0.
(5.13)
Interchanging X and Z for X, Z ∈ D in (5.13) and subtracting these relations, we obtain
φQ[φX, φZ] + Q[X, φZ] + h(X, φZ) − h(Z, φX) = 0
(5.14)
for any X, Z ∈ D and the assertion follows.
Now, we prove the following proposition.
Proposition 5.3. Let M be a CR-submanifold of a nearly trans-Sasakian manifold
M. Then
1
AφY Z − AφZ Y = α η(Y )Z − η(Z)Y + φP [Y , Z]
(5.15)
3
for any Y , Z ∈ D ⊥ .
Proof. For Y , Z ∈ D ⊥ and X ∈ T (M), we get
2g(AφZ Y , X) = 2g h(X, Y ), φZ
= g h(X, Y ), φZ + g h(X, Y ), φZ
= g X Y , φZ + g Y X, φZ
= g X Y + Y X, φZ
= −g φ X Y + Y X , Z
= −g Y φX + X φY + αη(X)Y + αη(Y )X − 2αg(X, Y )ξ, Z
= −g Y φX, Z − g X φY , Z − αη(X)g(Y , Z)
(5.16)
− αη(Y )g(X, Z) + 2αη(Z)g(X, Y )
= g Y Z, φX + g AφY Z, X − αη(X)g(Y , Z)
− αη(Y )g(X, Z) + 2αη(Z)g(X, Y ).
The above equation is true for all X ∈ T (M), therefore, transvecting the vector field
X both sides, we obtain
2AφZ Y = AφY Z − φY Z − αg(Y , Z)ξ − αη(Y )Z + 2αη(Z)Y
(5.17)
for any Y , Z ∈ D ⊥ . Interchanging the vector fields Y and Z, we get
2AφY Z = AφZ Y − φZ Y − αg(Y , Z)ξ − αη(Z)Y + 2αη(Y )Z.
(5.18)
Subtracting (5.17) and (5.18), we get
1
AφY Z − AφZ Y = α η(Y )Z − η(Z)Y + φP [Y , Z]
3
for any Y , Z ∈ D ⊥ , which completes the proof.
(5.19)
CR-SUBMANIFOLDS OF A NEARLY TRANS-SASAKIAN . . .
175
Theorem 5.4. Let M be a CR-submanifold of a nearly trans-Sasakian manifold M.
Then, the distribution D ⊥ is integrable if and only if
AφY Z − AφZ Y = α η(Y )Z − η(Z)Y ,
for any Y , Z ∈ D ⊥ .
(5.20)
Proof. First suppose that the distribution D ⊥ is integrable. Then [Y , Z] ∈ D ⊥ for
any Y , Z ∈ D ⊥ . Since P is a projection operator on D, so P [Y , Z] = 0. Thus from (5.15)
we get (5.20). Conversely, we suppose that (5.20) holds. Then using (5.15), we have
φP [Y , Z] = 0 for any Y , Z ∈ D ⊥ . Since rank φ = 2n. Therefore, either P [Y , Z] = 0 or
P [Y , Z] = kξ. But P [Y , Z] = kξ is not possible as P is a projection operator on D.
Thus, P [Y , Z] = 0, which is equivalent to [Y , Z] ∈ D ⊥ for any Y , Z ∈ D ⊥ and hence D ⊥
is integrable.
Corollary 5.5. Let M be a ξ-horizontal CR-submanifold of a nearly trans-Sasakian
manifold M. Then, the distribution D ⊥ is integrable if and only if
AφY Z − AφZ Y = 0
(5.21)
for any Y , Z ∈ D ⊥ .
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Falleh R. Al-Solamy: Department of Mathematics, Faculty of Science, King Abdul
Aziz University, P.O. Box 80015, Jeddah 21589, Saudi Arabia